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$ f(x)= x\ln x - x $

Wondering if my answer is right. Here is my process. I will simply find the derivative by using the product and difference rule.

$x \frac{d}{dx}[\ln x]+ \ln x\frac{d}{dx}[x]-\frac{d}{dx}[x]$

$x \cdot \frac{1}{x} + \ln x \cdot 1-1$

$1+ \ln x-1$

$y'=\ln x$

Is my answer correct?

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    $\begingroup$ Yes. Using the product rule, $f'(x)=1\cdot\ln(x)+x\cdot\frac{1}{x}-1=\ln(x)$. Just be careful of how it is written. $\endgroup$
    – user230715
    Commented Jul 1, 2015 at 13:04
  • $\begingroup$ You should include equal signs and specify that you are computing $f'(x)$. \begin{align*} f'(x) & = x \frac{d}{dx}[\ln x] + \ln x \frac{d}{dx}[x] - \frac{d}{dx}[x]\\ & = x \cdot \frac{1}{x} + \ln x \cdot 1 - 1\\ & = 1 + \ln x - 1\\ & = \ln x\end{align*} $\endgroup$ Commented Jul 1, 2015 at 13:23

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The answer is correct, but your notation is a bit off. Instead, write

\begin{align} f(x) &= x \ln (x) - x \\ \frac{\mathrm{d}}{\mathrm{d}x}f(x) &= \frac{\mathrm{d}}{\mathrm{d}x} (x \ln (x) - x) = x \frac{\mathrm{d}}{\mathrm{d}x} \ln(x) + \ln(x) \frac{\mathrm{d}}{\mathrm{d}x} x - \frac{\mathrm{d}}{\mathrm{d}x} x = \cdots \end{align}

Note the difference between $\frac{\mathrm{d}y}{\mathrm{d}x}$ in your case (incorrect) and $\frac{\mathrm{d}}{\mathrm{d}x}$ above (correct).

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  • $\begingroup$ Can you also solve this problem by following these steps: Steps in Logarithmic Differentiation: 1) Take the natural logarithms of both sides of an equation y=(x) and use the Laws of Logarithms to simplify\ 2) Differentiate implicitly with respect to x\ 3) Solve the resulting equation for y' . Why is this method useful? The books indicated this how we should solve these problems, but I found it confusing. $\endgroup$
    – Sunny
    Commented Jul 1, 2015 at 13:18
  • $\begingroup$ @Ritz Was the original post edited after you posted your answer? I don't see anything wrong with the notation now. $\endgroup$
    – Mark Viola
    Commented Jul 1, 2015 at 13:21
  • $\begingroup$ @Dr.MV It was indeed edited. $\endgroup$
    – Ritz
    Commented Jul 1, 2015 at 15:17
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Your solution is correct but your notation is inconsistent.

$\frac{dy}{dx}$ indicates the derivative of the function $y$ with respect to $x$; you can't write $\frac{dy}{dx}[x]$ as the latter would mean the derivative of $y$ times $x$.

You can either use $\frac d{dx} x$ (and so also $\frac d{dx} \ln x$) or simply $(\ln x)'$.

Note also that your original function is $f(x)$, so you're computing $\frac {d}{dx} f(x) = f'(x)$; so at the end you can't write $y'$ because $y$ was not defined in the first place

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  • $\begingroup$ Was the original post edited after you posted your answer? I don't see anything wrong with the notation now. $\endgroup$
    – Mark Viola
    Commented Jul 1, 2015 at 13:21
  • $\begingroup$ @Dr.MV Yes, many $y$'s have been removed, though there is still one at the end which is not supposed to be there :) $\endgroup$
    – Ant
    Commented Jul 1, 2015 at 13:22
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Your answer is correct but your notation inconsistent, as has been pointed out by other users. Another way of seeing that your answer is correct is to integrate $\ln x$.

$$\int \ln x \, \mathrm{d}x \stackrel{\text{IBP}}{=} x \ln x - \int \frac{x}{x} \, \mathrm{d}x = x \ln x - x + \mathrm{C}$$

Where the integration by parts is done by setting $u = \ln x$ and $\mathrm{d}v = 1$, a sneaky trick. Then $v = x$ and $x \, \mathrm{d}u = \mathrm{d}x$. The result follows.


I saw that you were looking for a way to find the derivative using logarithmic differentiation, this is done below:

$$\ln f(x) = \ln \left(x \ln x - x\right) = \ln \left(x(\ln x - 1)\right) = \ln x + \ln (\ln x - 1).$$

So differentiating implicitly with respect to $x$ yields $$\frac{f'(x)}{f(x)} = \frac{1}{x} + \frac{\frac{1}{x}}{\ln x - 1} = \frac{1}{x} + \frac{1}{x(\ln x -1)}$$

Multiplying through by $f(x)$ yields

$$f'(x) = \frac{x(\ln x - 1)}{x} + \frac{x(\ln x -1)}{x(\ln x - 1)} = \ln x -1 + 1$$

So we have $$f'(x) = \ln x.$$

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